Solution Chemistry Test Review Work Stations

Station 1 – Solubility Concepts

1. Identify the three factors that affect the rate of solution formation.

2. Identify the two components of a solution.3. What units do we use to express solubility?4. What are the factors that affect solubility? 5. What does a solution look like when it is

supersaturated? 6. How can you determine the saturation point of a

substance using lab techniques? (HINT: think about the procedures of the “Solubility Curve Lab” from last week)

Station 1 – Solubility Concepts Answer Key

• 1. Agitation, temperature change, particle size• 2. Solute, solvent• 3. grams solute/100 g of solvent• 4. temperature and pressure• 5. solid precipitates out when cooled• 6. raise the solution until all solute dissolves.

Lower temperature and record temperature at the instant solid crystals form in solution.

Station 2 – Solubility Curves

• See next slide for worksheet.

Solubility WS

DIRECTIONS: Use the solubility curve from the previous packet to answer the following questions.

___________1) What is the solubility of potassium nitrate in 100 grams of water at 20 °C ?

___________2) What is the solubility of potassium chloride in 100 grams of water at 50 °C ?

___________3 )What is the solubility of sodium chloride in 100 grams of water at 90 °C ?

___________4) What is the minimum temperature needed to dissolve 60 grams of potassium nitrate in 100 grams of water

___________5) What is the minimum temperature needed to dissolve 35 grams of potassium chloride in 100 grams of water

___________6) At what temperature do potassium chloride and potassium nitrate have the same solubility ?

___________7) If 110 grams of potassium chloride are mixed with 100 grams of water at 20 °C, how much will not dissolve ?

___________8) If 120 grams of potassium nitrate are mixed with 100 grams of water at 60 °C, how much will not dissolve ?

___________9) If 15 grams of potassium chloride are added to 100 grams of water at 30 °C, how much more must be added to saturate the solution?

___________10) If 85 grams of potassium nitrate are added to 100 grams of water at 70 °C, how much more must be added to saturate the solution?

___________11) 100 grams of water at 95 °C are saturated with sodium chloride. If this solution is cooled to 35°C, how much of the solid will precipitate?

___________12) 100 grams of water at 90 °C are saturated with potassium chloride. If this solution is cooled to 35°C, how much of the solid will precipitate?

___________13) How much potassium chlorate will dissolve in 200 grams of water at 70 °C ?

___________14) How much potassium nitrate will dissolve in 300 grams of water at 10 °C ?

___________15) How much potassium chloride will dissolve in 50 grams of water at 50 °C ?

Station 2 – Solubility Curves1) 34 g2) 43 g3) 39g4) 37 °C5) 25 °C 6) 20 °C 7) 76-77g8) 7 g9) 22g10) 51-55 g11) 2 g12) 18 g13) 70 g14) 66 g15) 21-22 g

Station 3 – Molarity

• 1. Define Molarity, then calculate the molarity of each solution:a) 1.0 mol KCl in 750 mL of solutionb) 0.5 mol MgCl2 in 1.5 L of solution

• 2. Calculate the moles and grams of solute in each solution:a) 1.0L of 0.50 M NaClb) 500 mL of 2.0 M KNO3

c) 250 mL of 0.10M CaCl2

Station 3 – Molarity

• 1. a.1.3 M KCl, b.0.33 M MgCl2

• 2. a. 0.5 mol NaCl, 29 g NaCl– b. 1.0 mol KNO3, 100 g KNO3

– C. 0.025 mol CaCl2, 2.8 g CaCl2

Station 4- molality

• 1. Distinguish between a 1 M solution and a 1 m solution.

2.Describe how you would prepare a 1.00 m solution of glucose. (i.e., grams of glucose needed in 1000 g of water?)

3.Describe how you would prepare a 0.500 m solution of sodium chloride. (i.e., grams of sodium chloride in 1000 g of water?)

Station 4- molality

• 1. 1 M solution: 1 mol of solute in 1 L of solution; 1 m solution: 1 mol of solute in 1000g/1 kg of solvent

• 2. Dissolve 180 g (1 mole) of glucose in 1 kg of water.

• 3. Dissolve 29.3 g of NaCl (0.5 mole) in 1 kg of water.

Station 5 – Percent Solutions

• 1. Calculate the grams of solute required to make the following solutions:a) 2500 g of saline solution (0.90% NaCl (m/m))b) 0.050 kg of 4.0% (m/m) MgCl2

2. What is the concentration ( in % (v/v)) of the following solutions?a) 25 mL of ethanol (C2H5OH) is diluted to a volume of

150 mL with water.b) 175 mL of isopropyl alcohol (C3H7OH) is diluted with

water to a total volume of 275 mL

Station 5 – Percent Solutions

• 1. a. 23 g NaCl B. 2.0 g MgCl2

2.a. 16% (v/v) ethanolb. 63.6 % (v/v) isopropyl alcohol

Station 6 – Molarity by Dilution

1. How many mL of 4.00 M KI are needed to prepare 250.0 mL of 0.760 M KI?

2. How could you prepare 250 mL of 0.20 M NaCl using only a solution of 1.0 M NaCl and water?

3. How many milliliters of 0.500 M KCl solution would you need to dilute to make 100.0 mL of 0.100 M KCl?

Station 6 – Molarity by Dilution

• 1. 47.5 mL• 2. Add 50 mL of the 1.0 M solution to a 250

mL volumetric flask, then fill to the mark.• 3. 20 mL.